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Pulley, two objects, frictionless plane; find acceleration and cable tension

  1. Jun 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A box of mass m1 is hanging from a cable of negligible mass, The cable rounds a pulley, where it is connected to a second box of mass m2 on a frictionless plane. The angle of the inclined plane is [tex]\theta[/tex] as measured from the flat ground. Both boxes are initially at rest. Find the acceleration of the boxes and the tension in the cable.


    2. Relevant equations



    3. The attempt at a solution

    Object A had mass m1 and object B has mass m2.

    I know that gravity and the cable tension are both affecting object A, and the cable tension, gravity, and a normal force from the plane are affecting object B. Therefore, I have:

    Object A: TA + FgA = m1a
    Object B: TB +FgB + n = m2a

    (where T, F, and a are all vectors)

    I tried drawing free-body diagrams, and I just get lost drawing all of the equations...I know I need to make component vectors, but there is a lot to the problem, and I'm feeling a little overwhelmed.

    Is this stuff correct so far? Thanks.
    Last edited: Jun 6, 2010
  2. jcsd
  3. Jun 6, 2010 #2

    Doc Al

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    What forces act on A? What forces act on B? (For B, all you care about are forces parallel to the incline.)
  4. Jun 6, 2010 #3
    Gravity and the cable tension are the forces acting on A.

    Gravity, the cable tension, and the normal force (the plane) act on force B.
  5. Jun 6, 2010 #4

    Doc Al

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    Good. Set up an equation with Newton's 2nd law.

    What are their components parallel to the incline? Set up an equation with Newton's 2nd law.

    You'll get two equations with two unknowns. Careful with signs. (If one block goes up, the other must go down.)
  6. Jun 6, 2010 #5
    Assuming A is heavier than B...

    for A, I get m1g - T = m1a

    The parallel forces for object B are Fgx and the T force... so I guess for B I get
    T-m2gsin[tex]\theta[/tex] = m2a.

    Am I ignoring the normal force because the object doesn't move in that direction so it doesn't affect the acceleration?

    If the above equations are right, then I can get the following equation for acceleration by making T= m2a + m2gsin(theta), and substituting it in for the T in object A's equation:

    a=m1g-m2gsin(theta) over m1+m2

    (sorry, formatting wouldn't work right for some reason)
  7. Jun 6, 2010 #6

    Doc Al

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    Right. You can consider the normal force if you like, but since the acceleration in that direction is zero, you'll just end up with N = m2gcosθ. You don't need it.

    There are several ways to combine the two equations; that way is perfectly fine. I assume you meant to write:
    a = (m1g - m2gsinθ)/(m1 + m2)

    Another way is simply to add the two equations. That way the T from one cancels the -T from the other.
  8. Jun 6, 2010 #7
    Yes, that it what I meant for the acceleration.

    To find the cable tension, I substituted acceleration back into one of the equations and got:

    T=m1g - m12g+m2gsinθ/m1+m2

    that looks really gross...is it still correct?

    And thank you so much for all your help.
  9. Jun 6, 2010 #8

    Doc Al

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    It looks a bit off to me, but it may be formatting again. In any case, your approach is correct. If you do it over more carefully, you can simplify it a bit so it doesn't look quite so ugly. (Hint: Express everything over m1 + m2.)
  10. Jun 6, 2010 #9
    What about this?

    T=(m1m2g(1+sinθ)) / (m1+m2)
  11. Jun 6, 2010 #10

    Doc Al

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    Looks good! :approve:
  12. Jun 6, 2010 #11
    Awesome. Thank you so, so much.
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